### 3.783 $$\int \frac{\sinh (x)}{1+\cosh (x)+\sinh (x)} \, dx$$

Optimal. Leaf size=18 $\frac{x}{2}-\frac{\sinh (x)}{2}+\frac{\cosh (x)}{2}$

[Out]

x/2 + Cosh[x]/2 - Sinh[x]/2

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Rubi [A]  time = 0.0253834, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {3131} $\frac{x}{2}-\frac{\sinh (x)}{2}+\frac{\cosh (x)}{2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[x]/(1 + Cosh[x] + Sinh[x]),x]

[Out]

x/2 + Cosh[x]/2 - Sinh[x]/2

Rule 3131

Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x
_)]), x_Symbol] :> Simp[((2*a*A - c*C)*x)/(2*a^2), x] + (-Simp[(C*Cos[d + e*x])/(2*a*e), x] + Simp[(c*C*Sin[d
+ e*x])/(2*a*b*e), x] + Simp[((-(a^2*C) + 2*a*c*A + b^2*C)*Log[RemoveContent[a + b*Cos[d + e*x] + c*Sin[d + e*
x], x]])/(2*a^2*b*e), x]) /; FreeQ[{a, b, c, d, e, A, C}, x] && EqQ[b^2 + c^2, 0]

Rubi steps

\begin{align*} \int \frac{\sinh (x)}{1+\cosh (x)+\sinh (x)} \, dx &=\frac{x}{2}+\frac{\cosh (x)}{2}-\frac{\sinh (x)}{2}\\ \end{align*}

Mathematica [A]  time = 0.039632, size = 18, normalized size = 1. $\frac{x}{2}-\frac{\sinh (x)}{2}+\frac{\cosh (x)}{2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[x]/(1 + Cosh[x] + Sinh[x]),x]

[Out]

x/2 + Cosh[x]/2 - Sinh[x]/2

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Maple [B]  time = 0.024, size = 28, normalized size = 1.6 \begin{align*} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}+{\frac{1}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)/(1+cosh(x)+sinh(x)),x)

[Out]

1/(tanh(1/2*x)+1)+1/2*ln(tanh(1/2*x)+1)-1/2*ln(tanh(1/2*x)-1)

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Maxima [A]  time = 1.16206, size = 14, normalized size = 0.78 \begin{align*} \frac{1}{2} \, x + \frac{1}{2} \, e^{\left (-x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(1+cosh(x)+sinh(x)),x, algorithm="maxima")

[Out]

1/2*x + 1/2*e^(-x)

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Fricas [A]  time = 2.438, size = 72, normalized size = 4. \begin{align*} \frac{x \cosh \left (x\right ) + x \sinh \left (x\right ) + 1}{2 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(1+cosh(x)+sinh(x)),x, algorithm="fricas")

[Out]

1/2*(x*cosh(x) + x*sinh(x) + 1)/(cosh(x) + sinh(x))

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Sympy [B]  time = 0.702367, size = 34, normalized size = 1.89 \begin{align*} \frac{x \tanh{\left (\frac{x}{2} \right )}}{2 \tanh{\left (\frac{x}{2} \right )} + 2} + \frac{x}{2 \tanh{\left (\frac{x}{2} \right )} + 2} + \frac{2}{2 \tanh{\left (\frac{x}{2} \right )} + 2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(1+cosh(x)+sinh(x)),x)

[Out]

x*tanh(x/2)/(2*tanh(x/2) + 2) + x/(2*tanh(x/2) + 2) + 2/(2*tanh(x/2) + 2)

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Giac [A]  time = 1.2229, size = 14, normalized size = 0.78 \begin{align*} \frac{1}{2} \, x + \frac{1}{2} \, e^{\left (-x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)/(1+cosh(x)+sinh(x)),x, algorithm="giac")

[Out]

1/2*x + 1/2*e^(-x)